CodeForces 495B Modular Equations

Modular Equations

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define i modulo j as the remainder of division of i by j and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form  in which a and b are two non-negative integers and x is a variable. We call a positive integer x for which  a solution of our equation.

Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.

Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers a and b determines how many answers the Modular Equation  has.

Input

In the only line of the input two space-separated integers a and b (0 ≤ a, b ≤ 10⁹) are given.

Output

If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .

Sample Input

Input

21 5

Output

2

Input

9435152 272

Output

282

Input

10 10

Output

infinity

题意就是给出一个方程 a%x=b，其中的a和b是已知的常数，问有多少个x的解，如果解有无穷多个就输出infinity

解题思路： t = a - b；如果t%x==0 中的x是一个解，那么t/x就是另外一个解当x>b && t/x>b，如果有一个不满足>b的条件就舍去。 那么就能以sqrt(t) 为一条界限去划分1-t，那么这一条界的左边是上述满足t%x==0的x的话，右边就是t/x，这样使得时间复杂度降低到sqrt(t)。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <queue>using namespace std;int main(){int i,j,k,t,m,n;int a,b;scanf("%d%d",&a,&b);if(a==b)printf("infinity\n");else if(b>a/2)printf("0\n");else{t =  a-b;int ans = 0;int tt = sqrt(t*1.0);for(i=1;i<=b;i++){if(t%i==0 && t/i>b)ans++;}for(i=b+1;i<=tt;i++){if(t%i==0 )ans+=2;}if((tt*1.0 == sqrt(t*1.0)) && tt>b)ans--;printf("%d\n",ans);}return 0;}